3.2.5 Transition metals

Question 1

Define transition metals

Answer 1

A metal that can form one or more stable ions with a partially filled d sub-level

Question 2

Which two period 4 d-block elements aren’t transition metals, why?

Answer 2

Scandium, as it only forms one ion, Sc³⁺, which has an empty d sub-level.

Zinc, as it only forms one ion, Zn²⁺, which has a full d sub-level.

Question 3

Which electrons are removed first when forming a positive transition metal ions?

Answer 3

Electrons from the s block, the electrons from the d block

Question 4

What are transition metals similar physical properties?

Answer 4

1. They all have a high density
2. They all have high melting and boiling points
3. Their ionic radii are approximately the same

Question 5

What chemical properties do transition metals possess?

Answer 5

1. They form complex ions
2. They form coloured ions
3. They’re good catalysts
4. They can exist in variable oxidation states

Question 6

What is a complex?

Answer 6

A central metal atom or ion surrounded by co-ordinately bonded ligands

Question 7

What is a ligand?

Answer 7

An atom or ion that donates a pair of electrons to a central transition metal ion to form a co-ordinate bond

Question 8

What is co-ordination number?

Answer 8

The number of co-ordinate bonds that are formed with the central metal ion

Question 9

What are the normal co-ordination numbers, why is this?

Answer 9

6, 4, and 2

If ligands are small (e.g. H₂O or NH₃) 6 can fit around the central metal ion. But if ligands are larger (e.g. Cl^-) only 4 can fit around the central metal ion

Question 10

What shape does co-ordinate number of 6 produce?
What is the bond angle?

Answer 10

Octahedral, bond angles are all 90^o

Question 11

What shape does a co-ordinate number of 4 produce?
What is the bond angle?

Answer 11

Tetrahedral, bond angles are all 109.5^o, however, can form a square planar shape, where bond angles are 90^o

Question 12

What shape does a co-ordinate number of 2 produce?
What is the bond angle?

Answer 12

Linear shape, bond angle 180^o

Question 13

What is the equation to work out the oxidation state of the metal ion

Answer 13

Oxidation state of the metal ion = Total oxidation state - Sum of the oxidation states of the ligands

Question 14

What is a monodentate ligand?

Answer 14

A ligand that can only form one co-ordinate bond

Question 15

What is a bidentate ligand?

Answer 15

A ligand that can can form two co-ordinate bonds

Question 16

What is a multidentate ligand

Answer 16

A ligand able to form two or more co-ordinate bonds

Question 17

Explain the structure of the haem group in haemoglobin?

Answer 17

• Haemoglobin contains Fe²⁺ ions, which are hexa-coordinated (six lone pair are donatated to them to form six co-ordinate bonds in an octohedral structure
• Four of the co-ordinate bonds come from a single multidentate ligand. Four nitrogen atoms from the same molecule co-ordinate bond around the Fe²⁺ to form a circle, aka haem
• The other two co-ordinate bonds come from a portien called globin, and either an oxygen or a water molecule - so the complex can transport oxygen to where it’s needed, and then swap it for a water molecule

Question 18

How does haemoglobin work?

Answer 18

• In the lung, where conc. of oxygen is high, an oxygen molecule substitutes the water ligand and bonds co-ordinately to the Fe(II) ion to form oxyhaemoglobin
• Oxyhaemoglobin is carried around the body
• When oxyhaemoglobon gets to a place where oxygen is needed, the oxygen molecule is exchanged for a water molecule
• The haemoglobin the return to the lungs to start the process again

Question 19

How does inhaling carbon monoxide disrupt haemoglobins ability to transport O₂?

Answer 19

• The haemoglobin swaps substitutes its water ligand for a carbon monoxide ligand, forming carboxyhaemoglobin
• Carbon monoxide is a strong ligand and doesn’t readily exchange with oxygen or water ligands
• Therefore haemoglobin cannot transport oxygen any more

Question 20

What are the symptoms of carbon monoxide poisoning?

Answer 20

• Headaches
• Dizziness
• Unconsciousness
• Death

Question 21

How can complex ions show optical isomerism?

Answer 21

• Complex ions show optical isomerism when an ion can exist in two forms that are non-superimposable mirror images
• This happens with octohedral complexes when three bidentate ligands co-ordinately bond with a central metal ion

Question 22

Explain how cis-trans isomers can be form in octohedral complex ions

Answer 22

Octohedral complexes with four monodentate ligands of one type and two monodentate ligands of another type can show cis-trans isomerism. If the odd ligands are opposite, it’s the trans isomer. If the they are next to each other it’s the cis isomer

Question 23

Explain how cis-trans isomers can be form in square planar complex ions

Answer 23

Square planar complex ions that have two pairs of ligands show cis-trans isomerism. When the paired ligands are opposite each other it’s the trans isomer, when they are next to eachother it’s the cis isomer

E.g. cis/transplatin

Question 24

Explain how ligands split the 3d sub-level into two energy levels

Answer 24

Normally the 3d orbitals of a transition metal all have the same energy, when a ligand bonds to the transition metal ion, some of the orbitals gain energy, splitting the 3d sub-level in two different energy levels
1. Electons tend to occupy the lower orbitals (ground state), to be promoted to the higher level they need equal energy to the energy gap, ΔE.
2. This energy come from visible light
3. The amount of energy needed to promote electrons depends on the central metal ion and it’s oxidation state, the ligands and the co-ordination number, as these effect the size of the energy gap, ΔE.

Question 25

Give the formula that can be used to determine the energy absorbed when electrons are promoted from the ground state to the excited state

Answer 25

ΔE=hv=hc/λ

• v= frequency of the light absorbed (Hz)
• h= plank’s constant (6.63x10^-34J s)
• c= the speed of light (3.00x10⁸ m s^-1)
• λ= wavelength of light absorbed (m)

Question 26

Why do complex ions appear coloured?

Answer 26

• When visible light hits a transition metal, some frequencies are absorbed when electrons are promoted to higher orbitals, the frequency absorbed depends of the energy gap (ΔE) - the larger the energy gap the higher the frequency absorbed
• The rest of the frequencies are transmitted or reflected and combine to produce complementary colour
• If there are no 3d electrons, or the 3d sub-level is full then no electrons will be promoted so no energy will be absorbed. Therefore the solution will appear white or colourless

Question 27

Name factors that effect the colour of complex ions (effect the energy gap)

Answer 27

• Changes in oxidation state
• Changes in co-ordination number
• Changes in ligand

Question 28

Give an two examples of how changes in oxidation state can change the colour of a complex ion

Answer 28

Complex: [Fe(H₂O)₆]²⁺_(aq) -> [Fe(H₂O)₆]³⁺_(aq)

Ox. state: 2+ -> 3+
Colour: Pale green -> Purple

AND…

Complex: [V(H₂O)₆]²⁺_(aq) -> [V(H₂O)₆]³⁺_(aq)

Ox state: +2 -> +3
Colour: violet -> green

Question 29

Give an example of how co-ordination number changes the colour of a complex ion

Answer 29

Complex: [Cu(H₂O)₆]²⁺ + 4Cl^- -> [CuCl₄]^2- + 6H₂O

Co-ordination number: 6 -> 4
Colour: blue -> yellow

Question 30

Give an example of how changing the ligand can effect the colour of a complex ion despite co-ordination number and oxidation state staying the same

Answer 30

[Co(H₂O)₆]²⁺ + 6NH₃ -> [Co(NH₃)₆]²⁺ + 6H₂O

Oxidation state: +2 -> +2
Colour: pink -> straw coloured

Question 31

Explain how optical spectroscopy can be used to determine the concentration of a solution of complex ions

Answer 31

1. White light is shone through a filter, which is chosen to only let the colour of light that is absorbed through the sample
2. The light passes through the sample to a colorimeter, which calculates how much light was absorbed by the sample
3. The more concentrated a coloured solution is the more light it will absorb. So you can use this measurement to work out the conc. of a solution of transition metal ions

Question 32

What needs to be done before using optical spectrocopy to calculate the conc. of an unknown solution?

Answer 32

A calibration curve needs to be produced, to produce a calibration curve a substance of a known conc. is measured and the results are plotted on a graph. One this has been done the graph can be used to determine the conc. of an unknown substance

Question 33

Give an example of a complete ligand substitution, were the ligands are of a similar size and are the same charge

Answer 33

The co-ordination number and shape of the complex ion won’t change

[Co(H₂O)₆]²⁺_(aq) + 6NH_3(aq) -> [Co(NH₃)₆]²⁺_(aq) + 6H₂O_(aq)

Colour goes from pink to straw coloured

Question 34

Give an example of a partial ligand substitution, were the ligands are of a similar size and are the same charge

Answer 34

The co-ordination number and shape of the complex ion won’t change

[Cu(H₂O)₆]²⁺_(aq) + 4NH_3(aq) -> [Cu(NH₃)₄(H₂O)₂]²⁺_{(aq) + 4H2O(l)}

Colour goes from blue to deep blue

Ammonia in excess*

Question 35

What happens in a substitution reaction if ligands are different sizes?

Give three examples using cobalt(II), copper(II) and iron(III)

Answer 35

There is a change of shape and co-ordination

[Co(H₂O)₆]²⁺_(aq) + 4Cl^-_(aq) <=> [CoCl₄]^2-_(aq) + 6H₂O_(l)
(Goes from pink to blue)

[Cu(H₂O)₆]²⁺_(aq) + 4Cl^-_(aq) <=> [CuCl₄]^2-_(aq) + 6H₂O_(l)
(Goes from pale blue to yellow)

[Fe(H₂O)₆]²⁺_(aq) + 4Cl^-_(aq) <=> [FeCl₄]^2-_(aq) + 6H₂O_(l)
(Goes from purple to yellow)

Question 36

What happens when you add sodium hydroxide or ammonia solution to a solution containing metal aqua ions?

Answer 36

A hydroxide precipitate is formed when a little bit of sodium hydroxide or ammonia solution is added

The hydroxide precipitate sometimes dissolves when excess NaOH or ammonia solution is added

Question 37

Why might a ligand substitution reaction not be reversible?

Answer 37

• If the new ligands form stronger bonds with the central metal ion than the previous ligand did, the change is less easily reversed - e.g. CN^- ions form stronger co-ordinate bonds with Fe³⁺ ions than H₂O molecules, so it’s hard to reverse this reaction:

[Fe(H₂O)₆]³⁺_(aq) + 6CN^-_(aq) -> [FeCN₆]^3-_(aq) + 6H₂O_(l)

• Multidentate/ bidentate ligands form more stable complexes than monodentate ligands, so a change like the one bellow is hard to reverse:

[Cu(H₂O)₆]²⁺_(aq) + 3NH₂CH₂CH₂NH_2(aq) -> [Cu(NH₂CH₂CH₂NH₂)₃]²⁺_(aq) + 6H₂O_(l)

Question 38

How is enthalpy effected in a ligand substitution reaction

Answer 38

• When a ligand exchange reaction occurs, co-ordinate bonds are broken and formed. The strength of the bonds being broken is often similar to the strength of the new bonds formed
• Therefore the enthalpy change for a ligand exchange reaction is usually small
• For example, the reaction substituting ammonia with ethane-1,2-diamine in a nickel complex has a very small enthalpy change of reaction:

[Ni(NH₃)₆]²⁺ + 3NH₂CH₂CH₂NH₂ -> [Ni(NH₂CH₂CH₂NH₂)₃]²⁺ + 6NH₃ ΔH = -13 kJ mol^-1

• This is actually a reversible reaction, but the equilibrium lies so far to the right that it is thought of as being irreversible
• [Ni(NH₂CH₂CH₂NH₂)₃]²⁺ is much more stable than [Ni(NH₃)₆]²⁺. This isn’t accounted for in enthalpy change

Question 39

What is the chelate effect

Answer 39

When monodentate ligands are substituted with bidentate or multidentate ligands, the number of particles in solution increases - the more particles the greater the entropy. Reactions that result in an increase in entropy are more likely to occur.

Question 40

Give the equation for when the hexadentate ligand EDTA^4- replaces monodentate ligands, and explain why the complex formed if more stable, and why it’s hard to reverse

Answer 40

[Cr(NH₃)₆]³⁺ + EDTA^4- -> [Cr(EDTA)]^- + 6NH³

2 particles -> 7 particles

Therefore enthalpy has increased, to reverse this reaction it would cause a large decrease in entropy, making it difficult to reverse

Question 41

What pneumonic can be used to remember the colour changes in vanadium from an oxidation state of +5 to +2

Answer 41

• YOU - Yellow - VO₂⁺_(aq) - +5
• BETTER - Blue - VO²⁺_(aq) - +4
• GET - Green - V³⁺_(aq) - +3
• VANADIUM - Violet - V²⁺_(aq) - +2

Question 42

Give the equation where vanadium(V) is reduced to vanadium(IV)

Answer 42

2VO₂⁺_(aq) + Zn_(s) + 4H⁺_(aq) -> 2VO²⁺_(aq) + Zn²⁺_(aq) + 2H₂O_(l)

Goes from yellow to blue

Question 43

Give the equation for when vanadium (IV) is reduced to vanadium(III)

Answer 43

2VO²⁺_(aq) + Zn_(s) + 4H⁺_(aq) -> 2V³⁺_(aq) + Zn²⁺_(aq) + 2H₂O_(l)

Goes from blue to green

Question 44

Give the equation for when vanadium(III) is reduced to vanadium(II)

Answer 44

2V³⁺_(aq) + Zn_(s) -> 2V²⁺_(aq) + Zn²⁺_(aq)

Goes from green to violet

Question 45

What is a redox potential and how does it tell you how easy it is to reduce an ion?

Answer 45

1. The redox potential of an ion tells you how easily it is reduced to a lower oxidation state - it’s the same as electrode potentials
2. The larger the redox potential, the less stable the ion will be, and so the more likely it is to be reduced.
3. Redox potentials are standard electrode potentials - they have been measured at a conc. of 1 mol dm³ against a standard hydrogen electrode, under standard conditions
4. The redox potential of an ion won’t always be the same as its standard electrode potential. It can vary depending on the enviorment that the ion is

Question 46

Explain how the redox potential of an ion isn’t always the same as its standard electrode potential, depending on it’s enviroment

Answer 46

• Ligands: standard electrode potentials are measured in aqueous solution, so any aqueous ions will be surrounded by water ligands. Different ligands may make the redox potential larger or smaller depending on how well they bind to the metal ion in a particular oxidation state
• pH: Some ions need H⁺ in order to be reduced, other release OH^- when reduced

For reactions such as these, the pH of the solution effects the size of the redox potential. In general, redox potentials will be larger in more acidic solutions, making the ion more easily reduced.

Question 47

Explain how tollens reagent contains a silver complex and how it’s charecteristics make it ideal for distinguishing between aldehydes and ketones

Answer 47

Silver is a transition metal that is most commonly found in the +1 oxidation state (Ag⁺). It’s easily reduced to silver metal:

Ag⁺_(aq) + e^- -> Ag_(s) Standard electrode potential = +0.80V

The standard electrode potential is large so Ag⁺ is easily reduced

Tollens reagent uses this reduction reaction to distinguish between aldehydes and ketones, as aldehydes can be oxidised to a carboxylic acid but ketones cannot

Question 48

Why are redox titrations used?

Answer 48

They allow to find the quantity of how much oxidising agent is needed to react with a quantity of reducing agent. If the conc. of either agent is known, you can use the titration results to work out the conc. of the other

Question 49

Why are transition metals able to be used for redox titrations?

Answer 49

Transition metals have variable oxidation states and therefore are often present in the oxidising or reducing agent. Thier colour changes also mean the end point can be spotted

Question 50

Give an example of a redox titration method

Answer 50

1. Measure out a quantity of reducing agent (e.g. Fe²⁺ ions or aqueous C₂O₄^2- ions) using a pipette, and put it in a conical flask
2. Using a measuring cylinder, add about 20cm³ of dilute sulfuric acid to the flask - this is an excess, therefore doesn’t need to be exact
3. Add the oxidising agent to the reducing agent using a burette, swirling the conical flask as you do so
4. The oxidising agent added reacts with the reducing agent. This reaction will continue until all of the reducing agent is used up. The next drop you add to the flask will give the mixture the colour of the oxidising agent ^1*
5. Stop when the mixture in the flask just becomes tainted with the colour of the oxidising agent (the end point) and record the volume of the oxidising agent added. This is the rough titration
6. Now do accurate titrations, do so until reading are within 0.10cm³ of each other (concordant results)

1* = A coloured reducing agent and a colourless oxidising agent instead, instead the colour would disappar at the end point
2* = This could also be done by adding the reducing agent to the oxidising agent
3* = Often aqueous potassium manganate(VII) which contains purple manganate(VII) ions, stong acidic conditions are required to reduce manganate(VII) ions

Question 51

Explain how before doing a redox titration, how to work out the reaction equation by balancing half equations

Use reduction of acidified manganate(VII) ions (MnO₄^-) by Fe²⁺ ions as an example

Answer 51

The half equation for the oxidation of iron is:

Fe²⁺_(aq) -> Fe³⁺_(aq) + e^-

For the other half of the equation:
1 - Manganate is being reduced

MnO₄^-_(aq) -> Mn²⁺_(aq)

2 - To balance the oxygens, water is added ot the other side of the equation

MnO₄^-_(aq) -> Mn²⁺_(aq) + 4H₂O_(l)

3 - Add H⁺ ions to the left side to balance the hydrogens

MnO₄^-_(aq) + 8H⁺_(aq) -> Mn²⁺_(aq) + 4H₂O_(l)

4 - Balance the charges by adding electrons

MnO₄^-_(aq) + 8H⁺_(aq) + 5e^- -> Mn²⁺_(aq) + 4H₂O_(l)

Now multiply the iron equation so the electons cancel out (x5)

Fe²⁺_(aq) -> Fe³⁺_(aq) + e^- -x5-> 5Fe²⁺_(aq) -> 5Fe³⁺_(aq) + 5e^-

Now combine both half equations to give the full redox equation

MnO₄^-_(aq) + 8H⁺_(aq) + 5Fe²⁺_(aq) -> Mn²⁺_(aq) + 5Fe³⁺_(aq) + 4H₂O_(l)

Question 52

Explain how you would calculate the conc. of a reagent using a titration using the example:

27.5 cm³ of 0.0200 mol dm^-3 aqueous potassium manganate(VII) reacted with 25.0 cm³ of acidified sodium ethandioate solution. Calculate the concentration of C₂O₄^2- ions in the solution.

2MnO₄^-_(aq) + 16H⁺_(aq) + 5C₂O₄^2-_(aq) -> 2Mn²⁺_(aq) + 8H₂O_(l) + 10CO_2(g)

Answer 52

1 - Work out the number of moles of MnO₄^- ions added to the flask.

_{Number of moles MnO4^- added = (concentration x volume(cm³))/1000 = (0.0200x27.5)/1000 = 5.50x10^-4 moles}

2 - Look at the balanced equation and work out how many moles of C₂O₄^2- react with every mole of MnO₄^-. Then you can work out the number of moles of C₂O₄^2- in the flask.

_{5 moles of C2O4^2- react with 2 moles of MnO4^-
So moles of C2O4^2- = (5.50 x 10^-4 x 5)/2 = 1.375 x 10^-3 moles}

3 - Work out the number of moles of C₂O₄^2- that would be in 1000cm³ (1 dm³) of solution - this is the concentration.

_{25.0 cm³ of solution contained 1.375 x 10^-3 moles of C2O4^2-.}

_{1000 cm³ of solution would contain ((1.375 x 10^-3) x 1000) / 25.0 = 0.0550 moles of C2O4^2-}

_{Therefore the concentration is 0.0550 mol dm^-3.}

Question 53

Why do transition metals work as good catalysts?

Give an example of a transition metal catalyst in a reaction

Answer 53

They can change oxidation states by gaining or losing electrons within their d orbitals, therefore they can transfer elctrons to increase the rate of reaction

E.g. The contact process:

V₂O₅ + SO₂ -> V₂O₄ + SO₃

_{Vanadium oxidises SO2 to SO, and is reduced itself}

V₂O₄ + 1/2 O₂ -> V₂O₅

_{The reduced catalyst is then oxidised by oxygen gas back to its original state}

Question 54

What is a heterogenous catalyst?

Answer 54

A catalyst that is in a different phase from the reactants

Question 55

Give two example of a heterogeneous catalyst

Answer 55

The haber process, Gasses are passed over a solid iron catalyst

The contact process for making sulfuric acid, vanadium oxide catalyst

Question 56

Give the equations for the contact process

Answer 56

V₂O₅ + SO₂ -> V₂O₄ + SO₃

V₂O₄ + 1/2 O₂ -> V₂O₅

Overall equation:

SO_2(g) + 1/2 O_2(g) –V₂O_5(s)–> SO_3(g)

Question 57

Give the overall equation for the Haber Process

Answer 57

N_2(g) + 3H_2(g) –Fe_(s)–> 2NH_3(g)

Question 58

Where does the reaction take place on a heterogeneous catalyst

Answer 58

The reaction happens on active sites on the surface of the heterogeneous catalyst

This means by increaseing the surface area of the catalyst there are more molecules that can react at the same time, increasing the rate of reaction

Question 59

Why are support mediums used on heterogeneous catalysts?

Answer 59

They make the surface area of the catalyst as large as possible. This helps to minimise the cost of the reaction, because only a small coating of catalyst is needed to provide a large surface area

Question 60

How can impurities poison heterogeneous catalysts, why is this bad?

Answer 60

1. Heterogeneous catalysts often work by adsorbing reactants onto active sites on their surface
2. Impurities in the reaction mixture may also bind to the catalyst’s surface surface and block reactants from being adsorbed
3. This reduces the surface area of the catalyst available to the reactants, slowing down the reaction
4. Catalyst poisoning increases the cost of a chemical process because less product can be made in a given time or with a certain amount of energy
5. One poisoned a catalyst may need replacing or regenerating, which costs money

Question 61

Explain how sulfer poisons the iron catalyst in the Haber Process

Answer 61

The hydrogen in the Haber Process is produced from methane. The methane is obtained from natural gas, which contains impurities, including sulfer compounds. Any sulfer that’s adsorbed onto the iron, forming iron sulfide, and stopping the iron from catalysing the reaction efficiently

Question 62

What are homgenous catalysts?

Answer 62

Catalysts that are in the same physical state as the reactants.

Usually a homogeneous catalyst is an aqueous catalyst for a reaction between two aqueous solutuions.

Question 63

How do homogeneous catalysts work?

Answer 63

By combining with the reactants to form an intermediate species which then reacts to form the products which reform the catalyst

Question 64

What does the enthalpy profile for a homogeneously catalysed reaction look like?

Answer 64

Has two humps in it, corresponding to the steps in the reaction.
The activation energy to from the intermediates is lower than needed to make the products directly from the reactants

Question 65

Explain why S₂O₈^2-_(aq) and 2I^-_(aq) react very slowly

Answer 65

The redox reaction between iodide ions and peroxodisulfate ions takes place very slowly as both ions are negatively charged. Therefore the ions repel eachother, so it’s likely theyy’ll collide and react

Question 66

Explain how Fe²⁺ catalyses S₂O₈^2-_(aq) and 2I^-_(aq)

Answer 66

When Fe²⁺ ions are added, the reaction speeds up because each stage of the reaction involves a positve and negative ion, therefore there is no repulsion.

Question 67

Give the steps involved (with equations) in Fe²⁺ catalysing S₂O₈^2-_(aq) and 2I^-_(aq)

Answer 67

1 - Fe²⁺ ions are oxidised to Fe³⁺ ions by the S₂O₈^2- ions

S₂O₈^2-_(aq) + 2Fe²⁺(aq) -> 2Fe³⁺(aq) + 2SO₄^2-_(aq)

2 - The newly formed intermediate Fe³⁺ ions now easily oxidise the I^- ions to iodine, and the catalyst is regenerated

2Fe³⁺_(aq) + 2I^-_(aq) -> I_2(aq) + 2Fe²⁺_(aq)

You can test for iodine by adding starch solution, if it goes black iodine is present

Question 68

What is auto catalysis?

Answer 68

When a product catalyses the reactants

Question 69

Explain how Mn²⁺(homogeneous catalyst) acts as a autocatalyst in the reaction between C₂O₄^2- and MnO₄^-

Answer 69

It’s an autocatalysis reaction because Mn²⁺ is a product of the reaction and acts as a catalyst for the reaction. This means as the reaction progresses and the amount of the product increases, the reaction speeds up (and eventually slows down as the amount of product decreases)

2MnO₄^-_(aq) + 16H⁺_(aq) + 5C₂O₄^2-_(aq) -> 2Mn²⁺_(aq) + 8H₂O_(l) + 10CO_2(g)

_{(reactant ions are both negative so repel each other causing an uncatalysed reaction to be very slow)}

1- Mn²⁺ catalyses the reaction by first reacting with MnO₄^- to form Mn³⁺ ions:

MnO₄^-_(aq) + 4Mn²⁺_(aq) + 4Mn²⁺_(aq) + 8H⁺_(aq) -> 5Mn³⁺_(aq) + 4H₂O_(l)

2 - The newly formed Mn³⁺ ions then react with C₂O₄^2-_(aq) to form carbon dioxide and reform the Mn²⁺ catalyst ion:

2Mn³⁺_(aq) + C₂O₄^2-_(aq) -> 2Mn²⁺_(aq) + 2CO_2(g)

Question 70

What prefix is given to a complex depending on the number of ligands it has?

Answer 70

2 = di
3 = tri
4 = tetra
5 = penta
6 = hexa

Question 71

What is a water ligand called when in a complex?

Answer 71

Water = aqua

Question 72

What is an ammonia ligand called when in a complex?

Answer 72

Ammonia = ammine

Question 73

What is a chloride ligand called when in a complex?

Answer 73

Chloride = chloro

Question 74

What is a hydroxide ligand called when in a complex?

Answer 74

Hydroxide = hydroxo

Question 75

What is a cyanide ligand called when in a complex?

Answer 75

Cyanide = cyano

Question 76

How does the name of the complex vary depending on whether it’s positve or negative

Answer 76

If the complex is positvely charged then it uses it’s normal name - e.g. heaxaquacopper(II)

If the complex is negatively charged then the suffic “ate” is added to the end, and sometimes it’s latin name is used instead - e.g. tetrachlorocuprate (despite the copper being positvely charged 1*)

1*The complex’s overall charge is negative however

Question 77

What is the name of chromium in an anionic complex?

Answer 77

Chromium = chromate

Question 78

What is the name of cobalt in an anionic complex?

Answer 78

Cobalt = colbaltate

Question 79

What is the name of copper in an anionic complex?

Answer 79

Copper = cuprate

Question 80

What is the name of iron in an anionic complex?

Answer 80

Iron = ferrate

Question 81

What is the name of manganese in an anionic complex?

Answer 81

Manganese = manganate

Question 82

What is the name of nickel in an anionic complex?

Answer 82

Nickel = nickelate

Question 83

What is the name of silver in an anionic complex?

Answer 83

Silver = argenate

Question 84

What is the name of vanadium in an anionic complex?

Answer 84

Vanadium = vanadate

Question 85

How is the oxidation state of the central metal ion displayed in the name of the complex ion?

Answer 85

In roman numerals in brackets

Question 86

In what order should ligands be named in a complex ion if there are more than two ligands present

Answer 86

In alphabetical order

The name of the ligand itself not it’s numbering prefix - e.g. diaqua

Question 87

In what order do you name a complex?

Answer 87

Number of ligands -> Names of ligands -> Metal ion -> Oxidation state in brackets